17 VECTOR CALCULUS

The Divergence Theorem VECTOR CALCULUS 17.9 The Divergence Theorem In this section, we will learn about: The Divergence Theorem for simple solid regions, and its applications in electric fields and fluid flow.

INTRODUCTION In Section 16.5, we rewrote Green’s Theorem in a vector version as: where C is the positively oriented boundary curve of the plane region D.

INTRODUCTION Equation 1 If we were seeking to extend this theorem to vector fields on , we might make the guess that where S is the boundary surface of the solid region E.

DIVERGENCE THEOREM It turns out that Equation 1 is true, under appropriate hypotheses, and is called the Divergence Theorem.

Notice its similarity to Green’s Theorem and Stokes’ Theorem in that: DIVERGENCE THEOREM Notice its similarity to Green’s Theorem and Stokes’ Theorem in that: It relates the integral of a derivative of a function (div F in this case) over a region to the integral of the original function F over the boundary of the region.

DIVERGENCE THEOREM At this stage, you may wish to review the various types of regions over which we were able to evaluate triple integrals in Section 15.6

We call such regions simple solid regions. We state and prove the Divergence Theorem for regions E that are simultaneously of types 1, 2, and 3. We call such regions simple solid regions. For instance, regions bounded by ellipsoids or rectangular boxes are simple solid regions.

The boundary of E is a closed surface. SIMPLE SOLID REGIONS The boundary of E is a closed surface. We use the convention, introduced in Section 16.7, that the positive orientation is outward. That is, the unit normal vector n is directed outward from E.

THE DIVERGENCE THEOREM Let: E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation. F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then,

THE DIVERGENCE THEOREM Thus, the Divergence Theorem states that: Under the given conditions, the flux of F across the boundary surface of E is equal to the triple integral of the divergence of F over E.

GAUSS’S THEOREM The Divergence Theorem is sometimes called Gauss’s Theorem after the great German mathematician Karl Friedrich Gauss (1777–1855). He discovered this theorem during his investigation of electrostatics.

OSTROGRADSKY’S THEOREM In Eastern Europe, it is known as Ostrogradsky’s Theorem after the Russian mathematician Mikhail Ostrogradsky (1801–1862). He published this result in 1826.

THE DIVERGENCE THEOREM Proof Let F = P i + Q j + R k Then, Hence,

THE DIVERGENCE THEOREM Proof If n is the unit outward normal of S, then the surface integral on the left side of the Divergence Theorem is:

THE DIVERGENCE THEOREM Proof—Eqns. 2-4 So, to prove the theorem, it suffices to prove these equations:

THE DIVERGENCE THEOREM Proof To prove Equation 4, we use the fact that E is a type 1 region: where D is the projection of E onto the xy-plane.

THE DIVERGENCE THEOREM Proof By Equation 6 in Section 15.6, we have:

THE DIVERGENCE THEOREM Proof—Equation 5 Thus, by the Fundamental Theorem of Calculus,

THE DIVERGENCE THEOREM Proof The boundary surface S consists of three pieces: Bottom surface S1 Top surface S2 Possibly a vertical surface S3, which lies above the boundary curve of D (It might happen that S3 doesn’t appear, as in the case of a sphere.)

THE DIVERGENCE THEOREM Proof Notice that, on S3, we have k ∙ n = 0, because k is vertical and n is horizontal. Thus,

THE DIVERGENCE THEOREM Proof—Equation 6 Thus, regardless of whether there is a vertical surface, we can write:

THE DIVERGENCE THEOREM Proof The equation of S2 is z = u2(x, y), (x, y) D, and the outward normal n points upward. So, from Equation 10 in Section 16.7 (with F replaced by R k), we have:

THE DIVERGENCE THEOREM Proof On S1, we have z = u1(x, y). However, here, n points downward. So, we multiply by –1:

THE DIVERGENCE THEOREM Proof Therefore, Equation 6 gives:

THE DIVERGENCE THEOREM Proof Comparison with Equation 5 shows that: Equations 2 and 3 are proved in a similar manner using the expressions for E as a type 2 or type 3 region, respectively.

THE DIVERGENCE THEOREM Notice that the method of proof of the Divergence Theorem is very similar to that of Green’s Theorem.

DIVERGENCE Example 1 Find the flux of the vector field F(x, y, z) = z i + y j + x k over the unit sphere x2 + y2 + z2 = 1 First, we compute the divergence of F:

DIVERGENCE Example 1 The unit sphere S is the boundary of the unit ball B given by: x2 + y2 + z2 ≤ 1 So, the Divergence Theorem gives the flux as:

Evaluate where: DIVERGENCE Example 2 F(x, y, z) = xy i + (y2 + exz2) j + sin(xy) k S is the surface of the region E bounded by the parabolic cylinder z = 1 – x2 and the planes z = 0, y = 0, y + z = 2

DIVERGENCE Example 2 It would be extremely difficult to evaluate the given surface integral directly. We would have to evaluate four surface integrals corresponding to the four pieces of S. Also, the divergence of F is much less complicated than F itself:

DIVERGENCE Example 2 So, we use the Divergence Theorem to transform the given surface integral into a triple integral. The easiest way to evaluate the triple integral is to express E as a type 3 region:

DIVERGENCE Example 2 Then, we have:

DIVERGENCE Example 2

UNIONS OF SIMPLE SOLID REGIONS The Divergence Theorem can also be proved for regions that are finite unions of simple solid regions. The procedure is similar to the one we used in Section 16.4 to extend Green’s Theorem.

UNIONS OF SIMPLE SOLID REGIONS For example, let’s consider the region E that lies between the closed surfaces S1 and S2, where S1 lies inside S2. Let n1 and n2 be outward normals of S1 and S2.

UNIONS OF SIMPLE SOLID REGIONS Then, the boundary surface of E is: S = S1 S2 Its normal n is given by: n = –n1 on S1 n = n2 on S2

UNIONS OF SIMPLE SOLID RGNS. Equation 7 Applying the Divergence Theorem to S, we get:

APPLICATIONS—ELECTRIC FIELD Let’s apply this to the electric field (Example 5 in Section 16.1): where S1 is a small sphere with radius a and center the origin.

APPLICATIONS—ELECTRIC FIELD You can verify that div E = 0 (Exercise 23). Thus, Equation 7 gives:

APPLICATIONS—ELECTRIC FIELD The point of this calculation is that we can compute the surface integral over S1 because S1 is a sphere.

APPLICATIONS—ELECTRIC FIELD The normal vector at x is x/|x|. Therefore. since the equation of S1 is |x| = a.

APPLICATIONS—ELECTRIC FIELD Thus, we have:

APPLICATIONS—ELECTRIC FIELD This shows that the electric flux of E is 4πεQ through any closed surface S2 that contains the origin. This is a special case of Gauss’s Law (Equation 11 in Section 16.7) for a single charge. The relationship between ε and ε0 is ε = 1/4πε0.

APPLICATIONS—FLUID FLOW Another application of the Divergence Theorem occurs in fluid flow. Let v(x, y, z) be the velocity field of a fluid with constant density ρ. Then, F = ρv is the rate of flow per unit area.

APPLICATIONS—FLUID FLOW Suppose: P0(x0, y0, z0) is a point in the fluid. Ba is a ball with center P0 and very small radius a. Then, div F(P) ≈ div F(P0) for all points in Ba since div F is continuous.

APPLICATIONS—FLUID FLOW We approximate the flux over the boundary sphere Sa as follows:

APPLICATIONS—FLUID FLOW Equation 8 This approximation becomes better as a → 0 and suggests that:

(This is the reason for the name divergence.) SOURCE AND SINK Equation 8 says that div F(P0) is the net rate of outward flux per unit volume at P0. (This is the reason for the name divergence.) If div F(P) > 0, the net flow is outward near P and P is called a source. If div F(P) < 0, the net flow is inward near P and P is called a sink.

SOURCE For this vector field, it appears that the vectors that end near P1 are shorter than the vectors that start near P1. Thus, the net flow is outward near P1. So, div F(P1) > 0 and P1 is a source.

Near P2, the incoming arrows are longer than the outgoing arrows. SINK Near P2, the incoming arrows are longer than the outgoing arrows. The net flow is inward. So, div F(P2) < 0 and P2 is a sink.

We can use the formula for F to confirm this impression. SOURCE AND SINK We can use the formula for F to confirm this impression. Since F = x2 i + y2 j, we have div F = 2x + 2y, which is positive when y > –x. So, the points above the line y = –x are sources and those below are sinks.